9709 MATHEMATICS P42 2014 GT

[aq0zsw1xde2c]

oh and for question 6. the deceleration was 67.5 and the value of R waa 15.5 N
the value of tension in the next part was 17.59 N

Yes that's exactly wat got but my acceleration was negative...

 

[Hadeelrm]

what are q7 answers?

 

[thementor]

Yes that's exactly wat got but my acceleration was negative...

yes of course it was.
Now tell me. did you get Tension = 17.59N or some other value?

 

[thementor]

what are q7 answers?

Q7 (I) TA = 0.25a +2.5, TB = 7.5-0.75a
(ii) a= 2 show
(iii) 1.2 ms^-1
(iv) -6 ms^-2 or simply the magnitude of the deceleration was 6

 

[raysonzaffar]

Q7 (I) TA = 0.25a +2.5, TB = 7.5-0.75a
(ii) a= 2 show
(iii) 1.2 ms^-1
(iv) -6 ms^-2 or simply the magnitude of the deceleration was 6

Do you remember the marks for each part of q7?
And the part iii could have been solved without finding the tensions, it was just kinematics right?

 

[A star]

Q7 (I) TA = 0.25a +2.5, TB = 7.5-0.75a
(ii) a= 2 show
(iii) 1.2 ms^-1
(iv) -6 ms^-2 or simply the magnitude of the deceleration was 6

what all my answers are correct cannot believe it XD XD XD XD A here i come so now i only loose 5 marks max thankyou

 

[Snowysangel]

Q7 (I) TA = 0.25a +2.5, TB = 7.5-0.75a
(ii) a= 2 show
(iii) 1.2 ms^-1
(iv) -6 ms^-2 or simply the magnitude of the deceleration was 6

Do you remember how you solved the last part? Also for the second last one, how'd you calculate the heights of A and B from the ground?

 

[aq0zsw1xde2c]

no i

yes of course it was.
Now tell me. did you get Tension = 17.59N or some other value?

t was the same... 17,59N

 

[raysonzaffar]

Do you remember how you solved the last part? Also for the second last one, how'd you calculate the heights of A and B from the ground?

The length of the string was 5.28m and the length of the table was 4m. So 5.25-4 = 1.28m
So the remaining 2 ssides each had a length of 1.28/2= 0.64
Hence the height of off the ground was 1-0.64=0.36

 

[thementor]

Do you remember the marks for each part of q7?
And the part iii could have been solved without finding the tensions, it was just kinematics right?

yes.
Part I and Part I each worht 3 marks. The last 2 parts were worth 2 marks each

 

[raysonzaffar]

yes.
Part I and Part I each worht 3 marks. The last 2 parts were worth 2 marks each

I got the acceleration as 2 by simultaneously solving the 2 equations where the tensions cancel. But I'm not sure whether my equation is right.

 

[thementor]

Do you remember how you solved the last part? Also for the second last one, how'd you calculate the heights of A and B from the ground?

yes. the total length of the string was 5.28
and the table's length was 4 so (5.28-4)/2 = 0.64. The height of each particle is 1-0.64=0.36 m
And for the last part. B has touched the ground. Tension in that string becomes 0
so 0-(0.25+2.5a +2 ) = 0.5a
so a = -6

 

[thementor]

what all my answers are correct cannot believe it XD XD XD XD A here i come so now i only loose 5 marks max thankyou

hahahaahah yayyyyyy. I am loosing more than you though. Do you remember the answer to Q2 (i) Did you get T=5???

 

[thementor]

I got the acceleration as 2 by simultaneously solving the 2 equations where the tensions cancel. But I'm not sure whether my equation is right.

If you got a=2 then it is correct.

 

[raysonzaffar]

hahahaahah yayyyyyy. I am loosing more than you though. Do you remember the answer to Q2 (i) Did you get T=5???

Yes t=5 seconds.

 

[A star]

yes. the total length of the string was 5.28
and the table's length was 4 so (5.28-4)/2 = 0.64. The height of each particle is 1-0.64=0.36 m
And for the last part. B has touched the ground. Tension in that string becomes 0
so 0-(0.25+2.5a +2 ) = 0.5a
so a = -6

i used equation of motion cuse the acceleration is constant

hahahaahah yayyyyyy. I am loosing more than you though. Do you remember the answer to Q2 (i) Did you get T=5???

can you write down the question idr it at all :/

 

[thementor]

Yes t=5 seconds.

Yesssssss. Finally.

 

[raysonzaffar]

If you got a=2 then it is correct.

So I'll only get marks for part ii not part i where i got the complete equation not just for Ta and Tb. Or will I get marks for (i) aswell?

 

[Snowysangel]

yes. the total length of the string was 5.28
and the table's length was 4 so (5.28-4)/2 = 0.64. The height of each particle is 1-0.64=0.36 m
And for the last part. B has touched the ground. Tension in that string becomes 0
so 0-(0.25+2.5a +2 ) = 0.5a
so a = -6

Oh now I remember...I though I was wring in assuming that they were both the same distance from the ground since it was me to ones nowhere in the question. I got -4 in the last question cause the only force that I took was friction and not the tension in the other string

 

[Hadeelrm]

was this exam harder or j13 was harder?